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Graph the linear equation and answer each of the following questions. y>2/3 x-2 What is the y-intercept? What is the slope (i

lozanna [386]

Answer:

(i)

y-intercept is -2

(ii)

slope is $\frac{2}{3}$

(iii)

slope is positive

(iv)

another point is (3,0)

Step-by-step explanation:

We are given equation as

$y=\frac{2}{3} x-2$

now, we can compare it with slope intercept form of line

y=mx+b

where

m is slope

b is y-intercept

we get

$m=\frac{2}{3}$

$b=-2$

So,

(i)

y-intercept is -2

(ii)

slope is $\frac{2}{3}$

(iii)

we can see that

$m=\frac{2}{3}$

is a fraction and positive

so, slope is positive

(iv)

we can select any random value of x and find y

that will give us second point

Let's assume

x=3

$y=\frac{2}{3}\times 3-2$

$y=2-2$

$y=0$

So, another point is (3,0)

(v)

Graph:

4 0

4 years ago

A major retail clothing store is interested in estimating the difference in mean monthly purchases by customers who use the stor

faust18 [17]

Answer:

Critical value is t = 1.9901

Step-by-step explanation:

We are given the results of the sampling :

In-House Credit Card National Credit Card Sample Size : 32 50

Mean Monthly Purchases : $45.67 $39.87

Standard Deviation : $10.90 $12.47

Also, the managers wished to test whether there is a statistical difference in the mean monthly purchases by customers using the two types of credit cards, using a significance level α of 0.05.

Firstly, we will specify our null and alternate hypothesis;

Let $\mu_1$ = Mean Monthly Purchases of In-House Credit Card

$\mu_2$ = Mean Monthly purchases of National Credit Card

So, Null Hypothesis, $H_0$ : $\mu_1-\mu_2$ = 0 {means that there is no difference in the mean monthly purchases by customers using the two types of credit cards}

Alternate Hypothesis, $H_0$ : $\mu_1-\mu_2\neq$ 0 {means that there is statistical difference in the mean monthly purchases by customers using the two types of credit cards}

The test statistics that will be used here is Two-sample t-test statistics;

T.S. = $\frac{(\bar X_1-\bar X_2)-(\mu_1-\mu_2)}{s_p \sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }$ ~ $t_n___1+n_2-_2$

where, $\bar X_1$ = Sample mean Purchases of In-House Credit Card = $45.67

$\bar X_2$ = Sample mean Purchases of National Credit Card = $39.87

$s_p$ = pooled variance

$n_1$ = In-house credit card sample = 32

$n_2$ = National credit card sample = 50

So, degree of freedom of t-value here is (32 + 50 - 2) = 80

Now, at 0.05 significance level, t table gives critical value of t = 1.9901 at 80 degree of freedom.

Therefore, the critical value assuming the population standard deviations are not known but that the populations are normally distributed with equal variances is t = 1.9901.

3 0

3 years ago

12c+12 =12c+9 How do I show the work for this

drek231 [11]

Answer:

hope that helps :))

Step-by-step explanation:

Step 1: Subtract 12c from both sides.

12c+12−12c=12c+9−12c

12=9

Step 2: Subtract 12 from both sides.

12−12=9−12

0=−3

6 0

3 years ago

Read 2 more answers

John has a deck of cards 52 cards of John remove the number to car from the deck what is the probability that he will pick up a

anastassius [24]

Answer:

3 out of 51

Step-by-step explanation:

If he removed one number two card from the deck then there should be three left since there are 4 of each number card in a deck so therefore there should be 51 cards in the deck but only three number two cards left in the deck

8 0

3 years ago

Help pleaseee !!!!!!!!!!!!

Nina [5.8K]

Girl with that what is you saying

3 0

3 years ago